The small part of the nucleus where ribosomal subunit formation takes place is called the nucleolus. This structure is found in eukaryotic cells and takes up about 25 percent of the volume of the nucleus. This structure is said to be the brain of the nucleus. It is made up of ribobucleic acids or the RNA and proteins. Its primary function is to transcript the ribosomal RNA (rRNA) and to combine these with the proteins resulting to the formation of an incomplete ribosome. Since it is closely related to a chromosomal matter of a cell and important in the production of ribosome, this structure is correlated to be one of the cause of a number of diseases.
<h3><u>Answer;</u></h3>
1) 2H2(g)+O2(g)⟶Δ2H2O(g)+483.6kJ - Exothermic
2) 6CO2(g)+6H2O(l)+15000kJ⟶C6H12O6(aq)+6O2(g) - Exothermic
3) 2Al(s)+Fe2O3(s)⟶Al2O3(s)+2Fe(s)+850kJ - Exothermic
<h3><u>Explanation</u>;</h3>
- Exothermic Reaction is a chemical reaction in which heat is given out is known as exothermic reaction, while, and endothermic Reaction is a chemical reaction in which heat energy is absorbed is known as endothermic reaction.
- The reaction between carbon dioxide and water is an example of an exothermic reaction. When most gases dissolve in solution, the process is exothermic meaning that heat is released as the gas dissolves
- The reduction of ferric oxide by aluminium is highly exothermic and therefore the iron formed will be in the molten state.
<span>We can solve this problem by assuming that the decay of
cyclopropane follows a 1st order rate of reaction. So that the
equation for decay follows the expression:</span>
A = Ao e^(- k t)
Where,
A = amount remaining at
time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560
M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>
The rate constant should
be given in the problem which I think you forgot to include. For the sake of
calculation, I will assume a rate constant which I found in other sources:
k = 5.29× 10^–4 s–1 (plug in the correct k value)
<span>Plugging in the values
in the 1st equation:</span>
A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )
A = 3.218 <span>× 10^–4 M (simplify
as necessary)</span>
Answer: 48,501 J/mol
Explanation:
1) Action barrier = activation energy = Ea
2) Data:
i) T₁ = 12°C = 12 + 273.15 K = 285.15K
ii) T₂ = 22°C = 22 + 273.15 K = 295.15 K
iii) rate constant = k: k₂ / k₁ = 2
iv) Ea = ?
3) Formula:
Arrhenius' law gives the relationship between the constant of reaction and the temperature:
4) Solution
By arranging the formula, you get:
㏑[k₂/k₁] =Ea/R [1/T₁ - 1/T₂]
Replace k₂ = 2k₁; T₁ = 285.15; and T₂ = 295.15
ln[2] = Ea/8.314 J/K mol × [1/285.15 - 1/295.15]K
Ea = ln [2] × 8.314 J/K mol / [1.18818×10⁻⁴K] = 48,501 J/mol
Answer:
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Explanation:
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